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Setler [38]
2 years ago
10

Need help on this ASAP !!!!

Chemistry
1 answer:
n200080 [17]2 years ago
7 0

Answer:

3 moles

Explanation:

This question is on stoichiometry of reactions

Given

2H₂ +O₂⇒2H₂O-------------------the equation is balanced

<u>Find the mole ratio</u>

2H₂ +O₂⇒2H₂O

2        1       2

O:H = 2:1---------------------------------every mole of hydrogen requires two moles of oxygen to complety react

Thus given 6 moles of hydrogen;

O:H

?:6

=6/2=3 moles

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Calculate q (in units of joules) when 1.850 g of water is heated from 22 °C to 33 °C. Report only the numerical portion of your
Minchanka [31]
When q is the heat energy in joules (J)

so, according to this formula, we can get q (in joule unit):

q = M*C*ΔT

when M is the mass of the water sample = 1.85 g

C is the specific heat capacity of water = 4.18 J/g.°C

and Δ T is the difference in temperature (Tf-Ti) = 33 - 22 = 11°C

So, by substitution, we will get the value of q ( in Joule):

∴ q = 1.85 g * 4.18 J/g.°C * 11 °C

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3 years ago
The freezing point of diet soda is higher than the freezing point of regular soda, but lower than 0 degrees celcius, the freezin
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Since soda Both regular or diet soda contains more solute than water , their freezing point is will consequently be lower than water

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The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
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Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

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Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

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Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

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